$\sin \alpha = \frac{3}{5} $ and $\cos \beta = -\frac{12}{13}$ . Find the values that $\cos(\alpha+\beta )$ can get. $\sin \alpha = \frac{3}{5} $ and $\cos \beta = -\frac{12}{13}$ . Find the values that $\cos(\alpha+\beta )$ can get.
Here $0<\alpha < \frac{\pi}{2}$ and $\frac{\pi}{2}<\beta<\pi$.
Yes I can find a value for $\cos(\alpha+\beta )$ by using $\cos(\alpha+\beta )=\cos \alpha \cos \beta - \sin \alpha \sin \beta$
But are there more values that  $\cos(\alpha+\beta )$ can get ?
 A: We have two well-known Pythagorean triples:
$$3^2+4^2=5^2 ,\qquad 5^2+12^2=13^2\tag{1}$$
hence $\sin\alpha=\frac{3}{5}$ and $0\leq\alpha\leq\frac{\pi}{2}$ implies $\cos\alpha=\frac{4}{5}$, as well as $\cos\beta=-\frac{12}{13}$ and $\frac{\pi}{2}\leq\beta\leq\pi$ implies $\sin\beta=\frac{5}{13}$. So, by the cosine sum formula:
$$ \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta = -\frac{48}{65}-\frac{15}{65}=\color{red}{-\frac{63}{65}.}\tag{2}$$
A: Since $\sin$ is a monotone increasing function on the interval $(0,\frac\pi2)$, and since $0<\frac35<1$, then we know, that there can only be one value of $\alpha$ in this interval for which $\sin(\alpha)=\frac35$, since $\sin$ will only take that value once.
Since $\cos$ is a monotone decreasing function on the interval $(0,\pi)$ and since $-1<-\frac{12}{13}<1$, we know that there can only be one value of $\beta$, since $\cos$ will only take that value once in the interval.
Thus, there can only be one $\cos(\alpha+\beta)$, since there can only be one unique $\alpha$ and $\beta$.
A: The formula for $\cos(\alpha + \beta)$ tells you how to compute this from $\cos \alpha, \sin \alpha, \cos \beta, \sin \beta$. You already know two of these numbers. The other two are not specified, but since $\cos^2 \gamma + \sin^2 \gamma = 1$ for any $\gamma$, there are not too many choices for each of these. And you know something about the ranges of $\alpha$ and $\beta$ which should help you narrow it down further. 
I hope you can take it from there.     
